Question

A 2000 kg car is speeding at 72 km/h. Determine the retarding force of brakes required to stop it in 300m on a level road.

Answer 1

Answer:

20 n

Explanation:

Given,

Mass of car, M=2000kg

Velocity, V=72×

18

5

=20ms

−1

Apply kinematic equation of motion

v

2

−u

2

=2as

0−20

2

=2a×20

a=−10ms

−2

Breaking force, F=ma=2000×10=20 kN

Apply first kinematic equation

v=u+at

t=

a

v−u

=

−10

0−20

=2 sec

Breaking force is 20 kNand time is 2sec.

Answer 2

Answer:

Force (F) applied by brakes is 4000 N

Explanation:

Gɪᴠᴇɴ :-

Mass (m) = 2000 kg

Initial velocity (u) = 72 ×5/18 = 20m/s

Final velocity (v) = 36×5/18 = 10 m/s

ᴛᴏ ғɪɴᴅ :-

Force (F) applied by brakes

sᴏʟᴜᴛɪᴏɴ :-

On using 1st equation of motion, we get,

➦ v = u + at

→ 10 = 20 + a × 5

→ 10 - 20 = 5a

→ 5a = -10

→ a = -10/5

→ a = -2 m/s²

Hence,

Acceleration (a) = 2 m/s²

Negative sign shows deceleration or retardation

Now,

We know that,

➦ Force = Mass × Acceleration

➦ F = ma

➦ F = 2000 × -2

➦ F = -4000 N

Hence,

Force (F) applied by brakes is 4000 N

Negative sign shows that force must be applied in opposite direction of motion .